B2-1 DNA Structure and Replication

Question 1

Parts of a Nucleotide

Answer 1

Five-carbon sugar – pentose
Nitrogenous base
Phosphate group

Question 2

Different types of nucleic acid, and difference in sugars

Answer 2

Deoxyribonucleic acid (DNA), in which the pentose sugar is deoxyribose. Deoxyribose-containing nucleotides, the deoxyribonucleotides, are the monomers of DNA

Ribonucleic acid (RNA), in which the pentose sugar is ribose. Ribose-containing nucleotides, the ribonucleotides, are the monomers of DNA

At the 2’ carbon of deoxyribose, the hydroxyl group (-OH) is replaced by a H atom

Small difference led to significant differences in structure and therefore functions of the 2 types of nucleic acids

Partial negative charge of the hydroxyl group in ribose repels the negative charge of phosphate, preventing the RNA chain from coiling in as tight a helix as it does in DNA
Hence, RNA is more susceptible to chemical and enzyme degradation

Question 3

Bonds in Nucleic Acids

Answer 3

Five-carbon sugars and occur as ring forms. In nucleic acids, the 5’ carbon is linked in an ester bond to the phosphate group and the 1’ carbon is linked in a glycosidic bond to the nitrogenous base

Question 4

Nitrogenous Base (Contains a structure, Types, DNA vs RNA)

Answer 4

Nitrogenous base has a nitrogen-containing ring structure

Nitrogenous bases fall into 2 types: purine and pyrimidine

Purine- 6-membered ring fused to a 5-membered ring – Adenine, Guanine
Pyrimidine –6-membered ring – Cytosine, Thymine, Uracil

DNA contains AGCT while RNA contains AGCU

Question 5

How nucleosides are formed

Answer 5

Pentose + Nitrogenous Base

Occurs with elimination of water – condensation reaction
1’ carbon of pentose is linked in a glycosidic bond to the nitrogenous base

2 types: Ribonucleoside and Deoxyribonucleoside

Question 6

How nucleotides are formed

Answer 6

Formed by further condensation between nucleoside and phosphate group, forming phosphoester bond between 5’ carbon of pentose and phosphate group

Number of phosphate groups linked to pentose sugar varies from 1 to 3

Question 7

Formation of di/poly nucleotides + Where the bond is + Characteristics of the bond

Answer 7

2 nucleotides join to form a dinucleotide by condensation between the 5’-phosphate group of 1 nucleotide and the 3’-hydroxyl group of the other to form a phosphodiester bond

Condensation reaction between nucleotides is repeated several million times to form a polynucleotide i.e. DNA or RNA

Phosphodiester bonds between 5’ phosphates and 3’ hydroxyl groups of nucleotides form a linear unbranched sugar-phosphate backbone

Phosphodiester bonds are strong covalent bonds, confer strength and stability on the polynucleotide chain. This is the basis in preventing breakage of the chain during DNA replication.

Question 8

Polarity / Directionality

Answer 8

Each DNA or RNA strand/chain has 2 free ends that are chemically different from each other

5’ end with a free 5’ carbon carrying a phosphate group

3’ end with a free 3’ carbon carrying a hydroxyl group

Every DNA and RNA molecule has an intrinsic polarity/Directionality. DNA or RNA base sequence is read in a 5’ to 3’ direction

Question 9

Main features of DNA

Answer 9

DNA consists of 2 polynucleotide strands/chains. Each strand forms a right-handed helix and the 2 strands coil around each other to form a double-helix.

1 DNA molecule = 1 DNA double-helix = 2 polynucleotide strands/chains

Diameter of the helix is uniformly 2nm. Hence, there is just enough space for 1 purine and 1 pyrimidine in the centre of the double helix

Strands run in opposite directions – antiparallel. One oriented in 5’ to 3’ while other is oriented in 3’ to 5’

Question 10

Sugar-phosphate backbone and why it’s good

Answer 10

Each strand has a sugar-phosphate backbone with

Phosphate groups that project outside the double helix since they are hydrophilic

Nitrogenous bases that orientate inwards towards the central axis at almost right angles

This arrangement is appealing because it puts relatively hydrophobic nitrogenous bases in the molecule interior and thus away from the surrounding aqueous medium

Question 11

CBP

Answer 11

Bases of opposite strands are bonded together by relatively weak hydrogen bonds

Specific cbp occurs between A & T (2 hydrogen bonds) and between C & G (3 hydrogen bonds)

Comes about this way because A-T and C-G pairs are the only ones that can fit the physical dimensions of the double helix. Moreover, they are in accord with Chargaff’s rules.

Significance of cbp:

Base sequence in one strand determines base sequence in the complementary strand

Weak hydrogen bonds make it relatively easy to separate the 2 strands of DNA by e.g. heating. A-T pair easier to separate by heating since A-T involves 2 hydrogen bonds and G-C 3

Question 12

Physical Properties of DNA - Distance, Grooves, etc

Answer 12

Base pairs are stacked 0.34nm apart along the central axis of the helix. As a result, hydrophobic interaction contributes to overall stability of the molecule

Double helix makes a complete turn every 10 base pairs, so each turn is 3.4nm

Grooves of unequal sizes between the sugar-phosphate backbones called major groove and minor groove. Both grooves are large enough to allow protein molecules to gain access and make contact with the bases

Question 13

Why form CBP

Answer 13

Steric restrictions

Sugar-phosphate backbone of each polynucleotide chain has a regular helical structure
DNA double-helix has a uniform diameter of 2nm
T & C are pyrimidines which have a single ring; A & G are purines, which are about twice as wide as pyrimidines
Solution is always to pair a purine with a pyrimidine

Hydrogen bond factors

Each nitrogenous base has chemical side groups such as H,N and O that can form hydrogen bonds with its appropriate partner
Such chemical side groups in purine and pyrimidines have well-defined positions
A is capable of forming 2 H bonds with T, while G is capable of forming 3 H bonds with C

Question 14

Important Notes on CBP

Significance + Variation of Linear Base Sequence

Answer 14

Significance

Since 3D structure of DNA is only stable when base pairs are complementary, this meant that the base sequence of 1 strand could determine the base sequence of its complementary strand
Necessary in DNA replication and transmission of genetic information stored

Variation of Linear Base Sequence

Although the base-pairing rules dictate the combinations of nitrogenous bases that form the ‘rungs’ of the double-helix, they do not restrict the base sequence along each DNA strand. Linear sequence of 4 bases can be varied in countless different ways

4^number of nucleotides = No. Of combinations of bases
Human beings have 3 x 10^9 nucleotide pairs therefore we have 4^3 x 10^9 combinations of bases
Hence, each gene has a unique base sequence

Question 15

Packing of DNA in Eukaryotic Chromosomes

Answer 15

A multilevel packing system that involves various proteins helping in the folding and condensation of DNA via a precise process is necessary to achieve a highly compact chromosome

Question 16

Stable, Invariant Storage of Genetic Info (Basic)

Answer 16

Genetic information that must be stored/preserved lies in the specific order of the base pairs i.e. base sequence must be stable and invariant

DNA allows for stable storage of genetic information – it is relatively resistant to spontaneous changes (mutations)

Question 17

Structural features that stabilise the DNA double-helix (a to d occur in both pro and euk)

Answer 17

Excessive hydrogen bonds between base pairs

Hydrophobic Interactions (or ‘stacking forces’) between the stacked base pairs

Exposure to outside influences of only the sugar-phosphate backbone

Nitrogenous bases being safely tucked inside the double-helix

Eukaryotes only: DNA double-helix being tightly wound around histones to form a repeating array of nucleosomes, which are eventually folded into higher order structures such as the chromosome, in which the DNA is prevented from thermal and physical damage

Question 18

Structural features that result in invariant base sequences

Answer 18

Specific, cbp between DNA strands. Hence,

Genetic information is redundant (present more than once) in the DNA molecule

If the base sequence in one of the 2 strands is accidentally altered, the cell discards the damaged strand. It then makes a perfectly good strand by using the remaining intact strand as a template, following Chargaff’s rules of cbp. The redundancy of genetic information helps to maintain its integrity.

Question 19

Significance of Base-pairing rule for replication

Answer 19

2 strands of DNA are complementary – each stores the information necessary to reconstruct the other

When a cell copies a DNA molecule, each strand serves as a template for ordering nucleotides into a new complementary strand

Where there was one double-stranded DNA molecule at the beginning of the process, there are now 2 – each an exact replica of the ‘parent’ molecule to ensure faithful transmission of genetic information

Question 20

Proposed model - Semi-conservative replication

Answer 20

1. The 2 DNA strands unwind and separate from each other, i.e. the hydrogen bonds between cbp are broken.
2. Each DNA strand then acts as a template for the assembly of a complementary strand.
3. Nucleotides line up singly along the template DNA strand according to cbp.
4. DNA polymerases join the nucleotides together

2 identical daughter DNA molecules are produced from a single parent DNA molecule
This model is described as semi-conservative: Each of the 2 daughter DNA molecules consists of 1 parental DNA strand and 1 newly-synthesised daughter DNA strand

Question 21

Conservative Model

Answer 21

Parental DNA molecule emerges from the replication process intact
It is conserved and generates DNA copies consisting of entirely new molecules

Question 22

Dispersive Model

Answer 22

All 4 strands of DNA following replication have a mixture of old and new DNA.

Question 23

The Experiment

Answer 23

1. For many generations, cells of the bacterium, E.coli were grown on medium containing only the ‘heavy’ isotope of nitrogen, 15N. 15N was incorporated into all the nitrogenous bases and the resulting DNA is known as ‘heavy DNA’.
2. The bacteria were then transferred to medium containing only the ‘light’ isotope of nitrogen, 14N and allowed to divide just once. This produces the 1st generation of bacteria.
3. Density-gradient centrifugation (in CsCl) was performed on a DNA extract from the bacteria. DNA is separated on the basis of density.
   Heavier 15N DNA molecules are denser than the 14N DNA molecules, hence 15N DNA molecules are spun further down the centrifuge tube where CsCl was denser.
   Lighter 14N DNA molecules were closer to the top where CsCl was less dense.
4. These bacteria were allowed to undergo a second round of replication and binary fission (in the presence of 14N) producing the 2nd generation. DNA was again separated by density centrifugation.

Question 24

CsCl Density-Gradient Centrifugation

Answer 24

DNA molecules move to the position where their density equals that of CsCl and ‘floats’ at that position.
Sedimentation of the CsC at the bottom of the spinning tube as a result of centrifugal forces.

Question 25

Results of DNA Experiment

Answer 25

Parental generation: All heavy DNA
1 gen: All hybrid DNA, intermediate in density
2 gen: 50% hybrid DNA, 50% light DNA

These results can only be achieved if replication is semi-conservative; 2 strands of parental heavy DNA separate and both strands act as a template for the assembly of a daughter light strand

Question 26

Mechanism of DNA Replication

The actual process is

Answer 26

Extremely complex. Timing of the steps is extremely precise – double helix must unwind and separate whilst the replication machinery copies the 2 antiparallel strands simultaneously

Extremely fast. Takes just a few hours to copy DNA in each human cell

Extremely accurate. Mutation rate is approximately 1 nucleotide change per 10^9 nucleotides each time DNA is replicated – Roughly the same for organisms as different as bacteria and humans

Requires cooperation of a large team of enzymes and other proteins, as well as the expenditure of ATP

Question 27

Location of oriR

Answer 27

DNA replication begins at one or more sites on the DNA molecule called oriR. Each of oriR is a specific sequence of nucleotides, which is generally A-T rich (there are only 2 hydrogen bonds between each A-T base pair, hence it is easier to disrupt the bonds as less energy is needed to overcome them).

Proteins that initiate DNA replication (initiator proteins) recognise this sequence and bind to the oriR sequence. DNA double-helix is separated into 2 strands, forming a replication ‘bubble’

Length of DNA unwound to initiate replication is typically ~50bp. ATP is required.

At each end of a replication bubble - Y-shaped structure called a replication fork, where the new strands of DNA are synthesised. The 2 replication forks move away from the oriR as replication proceeds bidirectionally (opp direction), until the entire DNA molecule is separated

Question 28

oriR - Prokaryotes

Answer 28

The prokaryotic chromosome is a small circular DNA molecule with a single oriR. Hence, DNA replication proceeds bidirectionally from the oriR to a termination site located approximately halfway around the circular chromosome, resulting in the synthesis of 2 daughter DNA molecules.

Question 29

oriR - Eukaryotes

Answer 29

Eukaryotic chromosome is much larger and consists of a linear DNA molecule, with multiple oriRs, hence multiple regions of the chromosome undergo replication at the same time.

Advantage of having multiple oriRs in eukaryotes is speed. Multiple replication bubbles form and eventually fuse, thus speeding up the copying of very long DNA molecules.

Replication takes approximately 8 hours in human cells with multiple oriRs. If it had only 1 oriR, it would take 100 times longer. This is important given the much larger size of a eukaryotic chromosome.

Replication begins at multiple oriRs, where the 2 parental strands separate to form replication bubble.

Bubbles expand laterally, as DNA replication proceeds bidirectionally.

Eventually, the replication bubbles fuse and synthesis of the daughter strands is complete.

Question 30

Helicases

Answer 30

After initiation, “unwinding” enzymes called helicases bind to one strand of the DNA strand

Using ATP as an energy source, helicases break the hydrogen bonds holding the 2 strands of DNA together. This unwinds the DNA double-helix and separates the parental DNA strands at the region of the replication fork.

Each of the 2 parental DNA strands serve as the template for the synthesis of a new DNA strand.

Question 31

Single-strand DNA-binding proteins

Answer 31

The unwound single-stranded portion of the DNA double helix is temporarily stabilised by the binding of SSB proteins.

This prevents the single-stranded (ss) DNA from re-annealing to reform the duplex, keeping the 2 parental strands in the appropriate single-stranded condition to act as template.

This also protects the ssDNA, which is very unstable, from being degraded by enzymes inside the cell

Question 32

Topoisomerases

Answer 32

Unwinding causes tighter twisting/supercoiling ahead of the replication fork, resulting in tension/torque

Topoisomerases cleave a strand of the helix to create a transient single-stranded nick (cut)

This relieves strain on the DNA molecule by allowing free rotation (swivelling) around the intact strand, then resealing of the broken strand

Question 33

RNA primer - Related limitation + Process

Answer 33

Limitation 1 of DNA polymerase

None of the DNA polymerases can initiate the synthesis of a DNA strand on its own. In other words, DNA synthesis cannot occur de novo.

Solution

To initiate synthesis of a DNA strand in the cellular context, an RNA primer is used.

A portion of the parental DNA strand serves as a template for making the RNA primer with the complementary base sequence.
An enzyme called primase joins the ribonucleotides to make the primer. The primer is about 10 nucleotides long in eukaryotes. Hydrolysis of ATP is involved.
The RNA primer provides a free 3’ OH end that DNA polymerase can extend, thereby priming the synthesis of the daughter DNA strand.
A DNA polymerase with 5’ to 3’ exonuclease activity later replaces (synthesis tbh, so it needs a 3’ group to connect to) the RNA nucleotides of the primers with DNA versions.

Question 34

CBP between templates and nucleotides

Answer 34

The parental DNA strands, separated at the replication fork and each primed with an RNA primer, serve as the templates for semi-conservative DNA replication.

DNA polymerase reads the template and assembles the nucleotide monomers (deoxyribonucleoside triphosphates, dNTPs) for the newly-synthesised daughter DNA strand based on CBP.

The error frequency of the newly-synthesised DNA strand must be low to ensure that replication is carried out accurately. When an incorrect base pair is recognised, DNA polymerase reverses its direction by one base pair of DNA. The 3’->5’ exonuclease activity (backwards) of the enzyme allows the incorrect base pair to be excised (proofreading).

Question 35

Phosphodiester bond formation between growing daughter DNA strand and incoming nucleotide

Answer 35

With an RNA primer anchoring the start of the daughter DNA strand, DNA polymerases catalyse the polymerisation of the strand

All DNA polymerases catalyse phosphodiester bond formation between a growing daughter DNA strand and an incoming nucleotide.

Because of active site specificity of the DNA polymerases, syntheses of both daughter DNA strand can only occur in one direction – 5 to 3. This is the other limitation of DNA polymerases

The addition of nucleotide (dNTP) to the growing DNA strand requires formation of a phosphoester bond between the free 3’ hydroxyl group of the last nucleotide in the growing strand and the free 5’ phosphate group of the incoming dNTP.

In this process, the incoming dNTP loses a pyrophosphate group (2 phosphate groups?) when they form the phosphoester bond with the growing daughter DNA strand. The energy released from pyrophosphate bond breakage is coupled to phosphoester bond formation.

Question 36

Limitation 2 of DNA Polymerase

Answer 36

DNA polymerases can only add dNTPs to the free 3’ end of a growing DNA strand, never to the 5’ end. Thus, a growing DNA strand can only elongate in the 5’ to 3’ direction.

Another factor to consider

The 2 strands of a DNA double-helix are antiparallel; sugar-phosphate backbones run in opposite directions

Problem

Continuous synthesis of both DNA strands at replication fork is not possible.

Question 37

Leading vs Lagging

Implication for Lagging Strand Synthesis

Answer 37

Leading Strand Synthesis

The complementary daughter DNA strand that is continuously synthesised as a single polymer along the template strand.

Polymerised in the mandatory 5’ to 3’ manner towards the replication fork

Lagging Strand Synthesis

The complementary DNA strand that is discontinuously synthesised as a series of short fragments known as Okazaki fragments.

Each Okazaki fragments is polymerised in the mandatory 5’ to 3’ manner against the overall direction of the replication fork. About 100 to 200 nucleotides per fragment in eukaryotes.

Question 38

End Replication Problem

Answer 38

Occurs in linear chromosomes as DNA polymerase is incapable of completely replicating all the way to the ends of linear chromosome, leading to shortening of telomeres with each round of DNA replication

More specifically, during DNA replication, the very end of the lagging strand is not replicated

Each time a cell with linear chromosome divides, a small section of the extreme 3’ end of the parental strand does not undergo DNA replication

This is because when the final RNA primer at the end of the lagging strand is removed, there is no upstream strand onto which DNA polymerase can build to fill the resulting gap

Hence, the daughter DNA strand resulting from lagging strand synthesis would be shortened with each round of replication

Question 39

Illustration of End Replication

Answer 39

The 5’ end of the daughter DNA strand resulting from lagging strand synthesis is not completely replicated as the replication fork reaches the end of a linear chromosome

Each daughter strand is leading at one end and lagging at the other

When the final primer of the terminal Okazaki fragment at the end of the lagging strand is removed by DNA polymerase, there is no upstream strand with a 3’ OH available to which DNA polymerase can add nucleotides to fill the resulting gap

After the first round of replication, the newly-synthesised daughter strands are shorter than its template strand